I believe the answer is A.
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>
My sample would be 4000 years old because on my graph, I had about 9 Virtualium left at trial 4 so I am guessing that it would be 4000 years old.
Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.